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Prove the Poisson formula for a general ball $B_R(x_0)\subset\mathbb{R}^n$ $$ u(x)=\frac{1}{\sigma_n R}\int_{S_R(x_0)}\frac{R^2-\lVert x-x_0\rVert^2}{\lVert\xi-x\rVert^n}\varphi(\xi)\, d\sigma\text{ for }x\in B_R(x_0) $$ by starting from the Poisson formula of the unit ball $B_1(0)\subset\mathbb{R}^n$ $$ u(x)=\frac{1}{\sigma_n}\int_{S_1(0)}\frac{1-\lVert x\rVert^2}{\lVert \xi-x\rVert^n}\varphi(\xi)\, d\sigma\text{ for }x\in B_1(0). (*) $$

Edit:

I do not come along with this, because I do not exactly know what to do resp. what to start with.

My first idea is, to consider the coordinate transformation

$$ \psi\colon\mathbb{R}^n\to\mathbb{R}^n, (x_1,x_2,\ldots,x_n)\longmapsto (Rx_1+x_1^0,\ldots,Rx_n+x_n^0) $$ Now, I would simply put that in (*), i.e.

$$ u(\psi(x))=\frac{1}{\sigma_n}\int_{S_1(0)}\frac{1-\lVert\psi(x)\rVert^2}{\lVert\xi-\psi(x)\rVert^n}\varphi(\xi)\, d\sigma $$

Now I have to integrate by substitution I think.

How can I do so? Do I first have to write that integral in n-dim. ball coordinates?

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    Surely this is just a simple (linear) change of variable in the integral? – Old John Nov 14 '13 at 22:37
  • I do not know but the work sheet says that this is to show. –  Nov 14 '13 at 22:43
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    If the worksheet says you are to deduce the general result form the special case of the unit ball, then I think my previous hint shows you the way. – Old John Nov 14 '13 at 22:45

1 Answers1

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You are substituting in the wrong place. $\varphi$ is defined on $S_R(x_0)$, so when using the Poisson integral for the unit ball, you must have $\varphi(\psi(\zeta))$ (with $\zeta\in S_1(0)$) in the integrand. And since that gives you a harmonic function in the unit ball, while you want a harmonic function in $B_R(x_0)$, the integral over the unit sphere gives you $u(\psi(y))$ for $y \in B_1(0)$, so

$$u(\psi(y)) = \frac{1}{\sigma_n} \int_{S_1(0)} \frac{1-\lVert y\rVert^2}{\lVert \zeta - y\rVert^n}\varphi(\psi(\zeta))\, d\sigma(\zeta).$$

Now writing $x = \psi(y)$ and $\xi = \psi(\zeta)$ gives you a relatively simple transformation to the desired form, since $\lVert \xi-x\rVert$ can be easily expressed using $\zeta-y$, and $\lVert x-x_0\rVert$ equally easily using $y$.

Daniel Fischer
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  • I understand how the formula for $u(\psi(y))$ arises. But now I do not understand how to continue with your transformations $x=\psi(y)$ and $\xi=\psi(\zeta)$. –  Nov 15 '13 at 14:15
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    $\lVert \xi-x\rVert^n = R^n\lVert\zeta-y\rVert^n$, $R^2-\lVert x-x_0\rVert^2 = R^2(1-\lVert y\rVert^2)$, $d\sigma(\xi) = R^{n-1}d\sigma(\zeta)$. Just rewrite and count factors of $R$. – Daniel Fischer Nov 15 '13 at 14:23
  • And one more question, sorry: Why $\lVert y\rVert^2$ in the integrand?.. and not $\lVert\psi(y)\rVert^2$? –  Nov 15 '13 at 14:24
  • The Poisson integral is the unit-ball integral of the function $u\circ\psi$. So if we evaluate that function in $y\in B_1(0)$, we plug in $y$, not $\psi(y)$. – Daniel Fischer Nov 15 '13 at 14:38
  • Hm, I do not understand this. Its really hard to me. –  Nov 15 '13 at 14:44
  • I think now I understood. –  Nov 15 '13 at 15:04
  • The professor said: Make a coordinate transformation and then use the transformation formula. Where did you use the transformation formula? –  Nov 15 '13 at 15:05
  • The transformation formula is $d\sigma(\psi(\zeta)) = R^{n-1}d\sigma(\zeta)$. The details of how to compute/determine that depend on how $d\sigma$ is defined in your course. – Daniel Fischer Nov 15 '13 at 15:11
  • the functional determinant is $R^n$, isn't it? So one has to use the transformation formula to get an expression with integral over $S_R(x_0)$? –  Nov 15 '13 at 15:29
  • I do not understand the step using the transformation formula. Damn ;( –  Nov 15 '13 at 15:33
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    Yes. The integral is over an $n-1$-dimensional hypersurface, so we don't need the determinant of the $\mathbb{R}^n\to\mathbb{R}^n$ map, but the $n-1$-dimensional version. Whichever way that was done in your course (there are many). – Daniel Fischer Nov 15 '13 at 15:35
  • Ok, maybe I am too stupid. But could you pls explain me how one exactly applies the transformation formula in order to come from $u(\psi(y))=\int_{S_1(0)} ... $ to the wanted integral form? –  Nov 15 '13 at 15:54
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    How exactly one applies the transformation formula depends on how you have defined $\int_{S_R(x_0)} H(\xi),d\sigma$. In some interpretation, you don't even use the transformation formula. If you view the integral as an integral over a rectangle $[0,2\pi]\times[0,\pi]^{n-2}\subset \mathbb{R}^{n-1}$ with a parametrisation of the sphere in spherical coordinates, you don't really use the transformation formula, you're just rewriting the integrands to see that both integrals are in fact the same. If you view the integral over the sphere as an integral with respect to an intrinsic measure on the – Daniel Fischer Nov 15 '13 at 16:29
  • involved spheres, you need to invoke the specific transformation formula for these measures (which turns out to just be a multiplication with the appropriate power of the radius). To use the standard transformation formula for integrals over subsets of $\mathbb{R}^k$ with respect to the Lebesgue measure, I must admit, I see no natural way, only artificial convoluted ways to obtain a change of the domain of integration. – Daniel Fischer Nov 15 '13 at 16:34
  • I mean that: http://de.wikipedia.org/wiki/Transformationssatz –  Nov 15 '13 at 16:36
  • How does one come to see $\Vert\xi-x\rVert^n=R^n\lvert\zeta-z)^n\rVert$ for example? Which transformation is behind that? And how does one come from integrating over $S_1(0)$ to integrating over $S_R(x_0)$? –  Nov 15 '13 at 16:45
  • And there we have the problem that the integral is over a lower-dimensional subset of $\mathbb{R}^n$. So this transformation formula does not apply. If we view the integral as an integral over a rectangle in $\mathbb{R}^{n-1}$ using the Gram determinant of the parametrisation of the sphere as weight, you don't use the transformation formula because both sphere are best parametrised over the same domain. – Daniel Fischer Nov 15 '13 at 16:46
  • let us continue this discussion in chat – Daniel Fischer Nov 15 '13 at 16:46